I got this really hard system as my extra credit problem, and I'm clueless on how to do it
Here is the problem

In the System 5x-2y=29 and kx+7ky=3000, k, x, y are all postive integers. Find the value of x+y

Thank you

2. do you know linear algebra?

or only solving for a variable and substituting?

3. We have to find the values of k, x, and y, but i dont know how to do it.

I Usually just try to eliminate or substitute in systems

4. I found two solutions (with the help of a small program I wrote)...

$(x,y,k)=(-1,-17,-25) \implies x+y=-18$
$~(x,y,k)=(19,33,12) \implies x+y=52$

The setup though, was fairly simple. First, you solve $5x-2y=29$ for $y$:

$y=\frac{5x-29}{2}$

Now you can create a table and find the $y$ value for $x=\{1,2,3,\ldots\}$. You will see that only the odd $x$ values will give integer $y$ values, and also that the $y$ values are increasing by 5 as $x$ increases by 2. So you have $\left( {x,y} \right) = \left[ \begin{gathered}
\vdots \\
- 3, - 23 \\
- 1, - 17 \\
1, - 12 \\
3, - 7 \\
\vdots \\
\end{gathered} \right]$

Now you solve the equation $kx+7ky=3000$ for $k$:

$k=\frac{3000}{x+7y}$

Now all that is left is to plug in your pairs of $x$ and $y$ and check to see if $k$ is an integer or not. It isn't the prettiest method, but it works. I do not know of a general algebraic method however, and I am pretty sure there is none. It seems like there should be an infinite amount of solutions (since there are three variables and only two equations), but the program only found those two for $x = \{ - 999999, - 999997, - 999995,\ldots,999999\}$

edit: I just noticed that you only need positive integers, but this all still stands. You have one solution, and apparently it is the only solution. Could anyone clarify why there aren't an infinite amount of solutions? Obviously the requirement that the variables be integers would have something to do with it (perhaps because they aren't dense in $\mathbb{R}$?), but I don't see why that would make the solution set finite.